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(1).The sum of first n terms of an AP is given by n (square)+8n.find its 12th term
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2
Sum of first n terms of the AP = n² + 8n
Sum of first 1 term = first term, a = 1² + 8×1 = 9
Sum of first 12 terms = 12² + 8×12 = 240
But we also know that sum of first n terms = (n/2)[a + an]
(an = nth term)
So sum of first 12 terms = (12/2)[9 + a₁₂]
⇒ 240 = 6×(9 + a₁₂)
⇒ 9 + a₁₂ = 240/6 = 40
⇒ a₁₂ = 40- 9
⇒ a₁₂ = 31
12th term is 31.
Sum of first 1 term = first term, a = 1² + 8×1 = 9
Sum of first 12 terms = 12² + 8×12 = 240
But we also know that sum of first n terms = (n/2)[a + an]
(an = nth term)
So sum of first 12 terms = (12/2)[9 + a₁₂]
⇒ 240 = 6×(9 + a₁₂)
⇒ 9 + a₁₂ = 240/6 = 40
⇒ a₁₂ = 40- 9
⇒ a₁₂ = 31
12th term is 31.
Answered by
2
Since ,
Sn = n² + 8n
Putting n = 1
S1 = 1 + 8 = 9
so first term is 9 i.e. a
Now putting n = 2
S2 = 4 + 8×2
S2 = 4 + 16 = 20
Now , s2-s1 = a2
a2= 20 - 9 = 11
so , d is a2-a1 = 2
so 12th term will be :-
a12 = a + 11d
a12 = 9 + 11×2
a12 = 9 + 22 = 31
Sn = n² + 8n
Putting n = 1
S1 = 1 + 8 = 9
so first term is 9 i.e. a
Now putting n = 2
S2 = 4 + 8×2
S2 = 4 + 16 = 20
Now , s2-s1 = a2
a2= 20 - 9 = 11
so , d is a2-a1 = 2
so 12th term will be :-
a12 = a + 11d
a12 = 9 + 11×2
a12 = 9 + 22 = 31
TPS:
S2-S1 = a2, not d
d = 2
a12 = 31
our maths teacher told that ans is 30 but u have 2 find
130
hmm... my teacher told that s2-s1 = d
Anyway, it's upto you... I am not going to argue. I gave a solution, it's upto you what you want to do.
ohkk.. bhaiya m confirm krta hu :) ,
umm.. u r ryt bhaiya s2-s1 = a2
(a1+a2) - a1 = a2
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